Q.A bag contains 5 red balls and 3 black balls. Two balls are drawn one after another without replacement. Find the probability that both balls drawn are red.

Since the draw is without replacement, the two draws are dependent — apply the general multiplication theorem, updating the bag's composition for the second draw.

Step 1 — Find the probability the first ball is red. The bag starts with $5$ red $+$ $3$ black $=8$ balls total. $P(\text{1st red}) = \dfrac{5}{8}$.

Step 2 — Find the conditional probability the second ball is also red, given the first was red. After removing one red ball (without replacement), the bag now has $4$ red and $3$ black $=7$ balls total. $P(\text{2nd red}\mid\text{1st red}) = \dfrac{4}{7}$.

Step 3 — Apply the multiplication theorem. Since the two draws are dependent (the bag's composition genuinely changes between draws), use the general form:

$P(\text{both red}) = P(\text{1st red})\times P(\text{2nd red}\mid\text{1st red}) = \dfrac{5}{8}\times\dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$ …

Treating 'without replacement' as if it were 'with replacement' — using $P(\text{2nd red})=5/8$ again instead of the correctly updated $4/7$ — is the single most common error in this topic; the composition of the bag gen …